# How do you find the amplitude, period, vertical and phase shift and graph y=3csc[1/2(theta+60)]-3.5?

##### 1 Answer
Aug 13, 2018

See explanation and graphical depiction.

#### Explanation:

csc values $\notin \left(- 1 , 1\right)$. $\theta$ is x, here..

$y = 3 \csc \left(\frac{1}{2} \left(x + \frac{\pi}{3}\right)\right) - 3.5$

$\notin \left(3 \left(- 1\right) - 3.5 , 3 \left(1\right) - 3.5\right) = \left(- 6.5 , - 0.5\right)$

$x \ne$ ( asymptotic zeros of $\sin \left(\frac{1}{2} \left(x + \frac{\pi}{3}\right)\right)$

$\left(2 k + 1\right) \frac{\pi}{3} ,$

k = 0, +-1, +-2, +-3, ...#

The period is the period of $\sin \left(\frac{1}{2} \left(x + \frac{\pi}{3}\right)\right) : \frac{2 \pi}{\frac{1}{2}} = 4 \pi$

Twisted $\infty$-wave amplitude : 1/2 ( period ) = $3 \pi$.

Vertical shift $= - 3.5 \Rightarrow$ axis is$y = - 3. 5$

Phase shift = $- \frac{1}{3} \pi$..

See graph, depicting all these aspects.
graph{(1/3(y+3.5) sin (1/2(x+1/3pi))-1)(x+7/3pi +0.0001y)(x-5/3pi +0.0001y)(y+3.5)(y+0.5) (y+6.5)(x + pi/3 +0.0001y)=0[ -10 10 -13.5 6.5]}

Here, the marked period : $x \in \left(- \frac{7}{3} \pi , \frac{5}{3} \pi\right)$

The phase shift marker: x = - pi/3